3.1.4 Sum/Product of the Roots

Define a polynomial by pn(x)=anxn+an1xn1+an2xn2++a1x1+a0=0p_n(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \dots + a_1 x^1 + a_0 = 0. The three most popular questions associated with the number sense test concerning roots of polynomials are: sum of the roots, sum of the roots taken two at a time, and product of the roots. For the polynomial pn(x)p_n(x) these values are defined:

  • Sum of the roots: an1an-\frac{a_{n-1}}{a_n}
  • Sum of the roots taken two at a time: an2an\frac{a_{n-2}}{a_n}
  • Product of the roots:
    • If nn is even: a0an\frac{a_0}{a_n}
    • If nn is odd: a0an-\frac{a_0}{a_n}

Let’s see what this means for our generic quadratics/cubics:

p2(x)=ax2+bx+c=0p_2(x) = ax^2 + bx + c = 0 and p3(x)=ax3+bx2+cx+d=0p_3(x) = ax^3 + bx^2 + cx + d = 0

p2(x)=ax2+bx+c=0p_2(x) = ax^2 + bx + c = 0

  • Sum of the roots: ba-\frac{b}{a}
  • Product of the roots: ca\frac{c}{a}

p3(x)=ax3+bx2+cx+d=0p_3(x) = ax^3 + bx^2 + cx + d = 0

  • Sum of the roots: ba-\frac{b}{a}
  • Product of the roots taken two at a time: ca\frac{c}{a}
  • Product of the roots: da-\frac{d}{a}

Since the quadratic only has two roots, the sum of the roots taken two at a time happens to be the product of the roots. You can extend the same procedure for polynomials of any degree, keeping in mind the alternating signs for the product of the roots. The following are practice problems:

Problem Set 3.1.4

The sum of the roots of 2x23x+1=02x^2 - 3x + 1 = 0 is
The sum of the roots of (x4)(x5)=0(x - 4)(x - 5) = 0 is
The sum of the roots of 3x32x2+x4=03x^3 - 2x^2 + x - 4 = 0 is
The product of the roots of x2+3x=7x^2 + 3x = 7 is
The sum of the roots of x29=0x^2 - 9 = 0 is
The sum of the roots of 4x2+3x=24x^2 + 3x = 2 is
The sum of the roots of (2x3)2=0(2x - 3)^2 = 0 is
The product of the roots of 5x38x2+2x+3=05x^3 - 8x^2 + 2x + 3 = 0 is
The product of the roots of 4x33x2+2x1=04x^3 - 3x^2 + 2x - 1 = 0 is
The sum of the roots of 3x3+2x2=93x^3 + 2x^2 = 9 is
The sum of the roots of x313x=12x^3 - 13x = 12 is
Let R, S, T be the roots of 2x3+4x=52x^3 + 4x = 5. Then R×S×T=R \times S \times T =
The product of the roots of 5x3+4x3=05x^3 + 4x - 3 = 0 is
The sum of the roots of (3x2)(2x+1)=0(3x - 2)(2x + 1) = 0 is
The sum of the product of the roots taken two at a time of 2x3+4x26x=82x^3 + 4x^2 - 6x = 8 is
The sum of the roots of 2x3+4x23x+5=02x^3 + 4x^2 - 3x + 5 = 0 is
The product of the roots of (2x1)(3x+2)(4x3)=0(2x - 1)(3x + 2)(4x - 3) = 0 is
Let R, S, T be the roots of 2x3+4x=52x^3 + 4x = 5. Then RS+RT+ST=RS + RT + ST =
The equation 2x3bx2+cx=d2x^3 - bx^2 + cx = d has roots r, s, t and rst=3.5rst=3.5, then d=d =
The sum of the roots of 3x2bx+c=03x^2 - bx + c = 0 is 12-12 then b=b =
If r, s, and t are the roots of the equation 2x34x2+6x=82x^3 - 4x^2 + 6x = 8 then rs+rt+st=rs + rt + st =
The sum of the roots of 4x3+3x22x1=04x^3 + 3x^2 - 2x - 1 = 0 is
The product of the roots of 4x33x2+2x+1=04x^3 - 3x^2 + 2x + 1 = 0 is
The sum of the roots of 5x3+4x3=05x^3 + 4x - 3 = 0 is
The equation 2x3bx2+cx=d2x^3 - bx^2 + cx = d has roots r, s, t. If r+s+t=2r + s + t = -2 then b=b =